# Series involving zeta functions

Let's have the following zeta function binomial: $\sum\limits_{n=1}^\infty \left(\frac{1}{n}-\frac{1}{n+1}\right)^2=\pi^2/3-3$.

Does anyone know the limit of the following zeta function binomial $\sum\limits_{n=1}^\infty \left(\frac{1}{n}-\frac{1}{n+1}\right)^4$?

-

Employ $(a-b)^4=a^4-4a^3b+6a^2b^2-4ab^3+b^4=a^4+6(ab)^2-4ab(a^2+b^2)$ on the following:

$$\blacksquare \; = \sum_{n=1}^{\infty} \left( \frac{1}{n} - \frac{1}{n+1} \right)^4= \sum_{n=1}^\infty \color{Red}{\frac{1}{n^4}}-\frac{4}{\color{Blue}{n^3(n+1)}}+\frac{6}{\color{Green}{n^2(n+1)^2}}-\frac{4}{\color{Blue}{n(n+1)^3}}+\color{Red}{\frac{1}{(n+1)^4}}$$

$$=\color{Red}{2\zeta(4)-1}+6\color{Green}{\sum_{n=1}^\infty \left(\frac{1}{n}-\frac{1}{n+1}\right)^2}-4\color{Blue}{\sum_{n=1}^\infty \left(\frac{1}{n}-\frac{1}{n+1}\right)\left(\frac{1}{n^2}+\frac{1}{(n+1)^2}\right)}$$

$$\frac{\pi^4}{45}+2\pi^2-19-4\color{Blue}{\blacktriangle}.$$

Note we used $\color{Green}{\bullet}=\pi^2/3-3$ and $\color{Red}{2\zeta(4)}=\pi^4/45$ above. Now we evaluate $\color{Blue}{\blacktriangle}$:

$$\color{Blue}{\blacktriangle}=\sum_{n=1}^\infty \frac{1}{n}\left(\color{Purple}{\frac{1}{n^2}}+\color{DarkOrange}{\frac{1}{(n+1)^2}}\right)-\frac{1}{n+1}\left(\color{DarkOrange}{\frac{1}{n^2}}+\color{Purple}{\frac{1}{(n+1)^2}}\right)$$

$$=\color{Purple}{\zeta(3)-[\zeta(3)-1]}+\color{DarkOrange}{\sum_{n=1}^\infty \frac{1}{n(n+1)}\left(\frac{1}{n+1}-\frac{1}{n}\right)}$$

$$=1-(\pi^2/3-3)=4-\pi^2/3.$$

Putting it all together, we obtain the final answer of

$$\blacksquare = \frac{\pi^4}{45}+\frac{10\pi^2}{3}-35.$$

The two general tools in my answer to this and your previous zeta question are:

• $\displaystyle \frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}$ to either break or put together terms, and
• $a^nb^m+a^mb^n=(ab)^k(a^l+b^l)$, where $k=\min\{n,m\}$ and $l+k=\max\{n,m\}$.

The first breaks down terms and the second is just for some (perhaps strained) efficiency. It only takes these formulas (plus $\zeta(2s)$'s) to compute $\sum_{n=1}^\infty (n^2+n)^{-k}$ for any $k$, and any $\zeta(2s+1)$'s that appear will always cancel each other out (up to a rational number).

-
@ Anon Only your answer is accepted. –  Vassilis Parassidis Feb 18 '12 at 5:41

$$\sum_{n=1}^{\infty} \left( \frac{1}{n} - \frac{1}{n+1} \right)^4 = \frac{ \pi^2 (\pi^2 + 150) }{45} - 35.$$

This is derived in the same manner as to the other sum you gave - expand the binomial, notice the obvious telescoping series, and sum the others using what is known about $\zeta (2n).$

-
@Ragib.Can you please present the detailed calculations. –  Vassilis Parassidis Feb 18 '12 at 4:12